Section 1: Numbers, Variables, Functions, and Equations

Calculus Lecture Notes

Learning Objectives

By the end of this 90-minute lecture, you will be able to:

1. Understand numbers as mathematical abstractions and variables as placeholders
2. Recognize functions as mathematical abstractions of relationships between variables
3. Present functions using text, tables, graphs, code, mathematical formulas
4. Functions of multiple variables
5. Apply functions to describe real-world relationships
6. Solve equations by finding roots

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Introduction: Mathematics as a languange for abstract objects

Mathematics is often called the “universal language” because it provides precise tools for describing patterns, relationships, and changes in our world. Today, we begin our journey into calculus by exploring its fundamental building blocks: numbers, variables, functions, and equations.

Think of this section as learning the alphabet before writing poetry. Just as poets use words to express complex emotions and ideas, mathematicians use functions to express complex relationships and changes.

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Knowledge Point 1: Numbers as Mathematical Abstractions and Variables as Placeholders

The Power of Abstract Thinking

Abstract Numbers Concept

In the real world, we can observe two apples and three potatoes. However, we cannot observe “2” and “3” themselves. Yet everyone knows what “2” means and what “3” means, despite the fact that we never directly observe these abstract concepts.

This remarkable ability reveals something profound: everyone has the capacity to think abstractly! This is the fundamental capacity that makes mathematics possible. When you recognize that two apples and two bananas share the same “twoness,” you are already thinking like a mathematician. This abstraction—moving from concrete objects to abstract concepts—is the foundation of all mathematical reasoning.

Verbal Scenario: The Coffee Shop Story

Imagine you’re managing a coffee shop. Every day, you observe patterns: - The number of customers varies throughout the day - Your revenue depends on how many cups you sell

These observations involve numbers (like 50 customers, $300 revenue) and relationships between quantities. Mathematics gives us tools to work with these relations precisely.

Formal Presentation

Definition 1.1 (Numbers as Abstractions): Numbers are mathematical objects that represent quantities, measurements, or positions. They abstract away specific physical properties to focus on quantitative relationships.

Definition 1.2 (Variables): A variable is a symbol (usually a letter) that represents an unknown or changing quantity. Variables serve as placeholders that can take on different values.

Types of Numbers in Calculus:

• Natural numbers (ℕ): 1, 2, 3, 4, … where \(1, 2, 3, 4 \in \mathbb{N}\)
• Integers (ℤ): …, -2, -1, 0, 1, 2, … where \(-2, -1, 0, 1, 2 \in \mathbb{Z}\)
• Rational numbers (ℚ): fractions like 1/2, -3/4 where \(\frac{1}{2}, -\frac{3}{4} \in \mathbb{Q}\)
• Real numbers (ℝ): all numbers on the number line, including √2, π, e where \(\sqrt{2}, \pi, e \in \mathbb{R}\)
• Complex numbers (\(\mathbb{I}\)): A complex number is a number that has both a real part and an imaginary part. It is written in the standard form:\(a+ib\), where \(a\) is the real part and \(ib\) is the imaginary part, and \(i\) is the imaginary unit with \(i^2=-1\).

Important Relationship: \(\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{I}\)

This means every natural number is also an integer, every integer is also a rational number, and every rational number is also a real number.

Graphical Presentation of Real Numbers

# Create a number line visualization
x <- seq(-5, 5, by = 0.1)
y <- rep(0, length(x))

# Create the plot
plot(x, y, type = "l", lwd = 3, col = "black", 
     xlim = c(-5, 5), ylim = c(-0.5, 0.5),
     xlab = "Real Numbers", ylab = "", yaxt = "n",
     main = "The Real Number Line")

# Add tick marks and labels
points(c(-4, -3, -2, -1, 0, 1, 2, 3, 4), rep(0, 9), pch = 19, cex = 1.2)
text(c(-4, -3, -2, -1, 0, 1, 2, 3, 4), rep(-0.2, 9), 
     labels = c(-4, -3, -2, -1, 0, 1, 2, 3, 4))

# Highlight special numbers
points(pi, 0, pch = 19, col = "red", cex = 1.5)
text(pi, 0.3, "π ≈ 3.14", col = "red", cex = 0.8)

points(-sqrt(2), 0, pch = 19, col = "blue", cex = 1.5)
text(-sqrt(2), 0.3, "-√2 ≈ -1.41", col = "blue", cex = 0.8)

# Add arrows
arrows(-5.5, 0, -5, 0, length = 0.1, lwd = 2)
arrows(5, 0, 5.5, 0, length = 0.1, lwd = 2)

The Real Number Line showing different types of numbers

Variable as a placeholder for numbers

Key Learning Points:

• Variables in Action: See how changing the input variables (customers, price) immediately affects the output (revenue)

• Relationship: Revenue = customers × price demonstrates a constant relationship, regardless of how the values of the variables change.

Interactive Quiz 1.1

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Knowledge Point 2: Functions as Mathematical Notion of Relations

Verbal Scenarios

1.1 The Coffee Shop Story (Revisited)

Let’s return to our coffee shop example with a mathematical perspective. Suppose you’ve established a fixed price of $3.50 per cup (prices don’t change very often in real businesses, so we treat price as a parameter). Your daily revenue depends linearly on the number of cups sold:

Revenue Function: R(Q) = 3.50Q

Where: - Q = number of cups sold (independent variable) - R = total revenue in dollars (dependent variable)
- 3.50 = price per cup (parameter)

This is a linear function because revenue increases at a constant rate with each additional cup sold.

1.2 The Delivery Service Story

Consider a delivery service with a tiered pricing structure: - Short distances (0-5 km): $10 base fee - Medium distances (5-15 km): $10 + $2 per additional km beyond 5 km - Long distances (15+ km): $10 + $20 + $1.50 per additional km beyond 15 km

Piecewise Function: \[C(d) = \begin{cases} 10 & \text{if } 0 \leq d \leq 5 \\ 10 + 2(d-5) & \text{if } 5 < d \leq 15 \\ 10 + 20 + 1.5(d-15) & \text{if } d > 15 \end{cases}\]

This is a piece-wise linear function with different slopes for different distance ranges.

Formal Definition

Definition 2.1 (Function): A function is a rule that assigns to each input value exactly one output value. We write f(x) = y, where x is the input (independent variable) and y is the output (dependent variable). \(f: \mathbb R \to \mathbb R\)

Key Properties:

1. Domain: The set of all possible input values

2. Range: The set of all possible output values

3. One-to-one correspondence: Each input produces exactly one output

For a real function: \(f: \mathbb R \to \mathbb R\)

Function Presentation Functions can be represented in multiple equivalent ways:

1. Tabular Representation: Discrete input-output pairs

2. Graphical Representation: Visual plot of the relationship

3. Algebraic Representation: Mathematical formula

4. Computational Representation: Algorithm or code

1.3 The Weather Station Story

A weather station records temperature throughout the day. The same data can be presented in multiple ways: - Table: Hour-by-hour temperature readings - Graph: A smooth curve showing temperature changes - Formula: A mathematical equation that predicts temperature - Code: An algorithm that computes temperature values

Each representation reveals different aspects of the relationship and serves different purposes. For example, a table gives exact values at specific times, a graph shows overall patterns and trends, a formula allows calculation of temperature at any time, and code enables efficient computation for large datasets.

Multiple Representations: The Weather Station Example

# Create a comprehensive example showing all representations
set.seed(123)

# Define a temperature function (simplified model)
temperature_func <- function(hour) {
  # Base temperature with daily variation
  base_temp <- 20
  daily_variation <- 8 * sin(2 * pi * (hour - 6) / 24)
  noise <- rnorm(length(hour), 0, 1)
  return(base_temp + daily_variation + noise)
}

# Generate data
hours <- 0:23
temps <- temperature_func(hours)

# Create a 2x2 layout
par(mfrow = c(2, 2), mar = c(4, 4, 3, 2))

# 1. Table representation (as a plot)
plot(1, type = "n", xlim = c(0, 10), ylim = c(0, 10), 
     xlab = "", ylab = "", main = "Tabular Representation", axes = FALSE)
text(5, 9, "Hour | Temperature (°C)", cex = 1.2, font = 2)
for(i in 1:8) {
  text(5, 8.5 - i*0.8, paste(hours[i], "|", round(temps[i], 1)), cex = 1)
}
text(5, 1, "... (continues for 24 hours)", cex = 0.9, style = 3)

# 2. Graphical representation
plot(hours, temps, type = "b", pch = 19, col = "blue", lwd = 2,
     main = "Graphical Representation", 
     xlab = "Hour of Day", ylab = "Temperature (°C)")
grid()

# 3. Smooth function overlay
hours_smooth <- seq(0, 23, by = 0.1)
temps_smooth <- 20 + 8 * sin(2 * pi * (hours_smooth - 6) / 24)
plot(hours_smooth, temps_smooth, type = "l", lwd = 3, col = "red",
     main = "Algebraic Representation", 
     xlab = "Hour of Day", ylab = "Temperature (°C)")
text(12, 25, "T(h) = 20 + 8sin(2π(h-6)/24)", cex = 1.1, font = 2)
grid()

# 4. Code representation
plot(1, type = "n", xlim = c(0, 10), ylim = c(0, 10), 
     xlab = "", ylab = "", main = "Computational Representation", axes = FALSE)
code_lines <- c(
  "temperature <- function(hour) {",
  "  base_temp <- 20",
  "  variation <- 8 * sin(2*pi*(hour-6)/24)",
  "  return(base_temp + variation)",
  "}"
)
for(i in 1:length(code_lines)) {
  text(0.5, 9 - i*1.2, code_lines[i], cex = 0.9, adj = 0, family = "mono")
}

Multiple Representations of the Same Function

par(mfrow = c(1, 1))  # Reset layout

Common Function Types

Explore each function type interactively below. Click the buttons to show/hide graphs and see how different functions capture different relationships:

Graphical Presentation: Comparing All Function Types

The following graph shows all six common function types over the interval [-4, 8], demonstrating how different functions describe different relationships between dependent and independent variables:

# Create a comprehensive comparison of all function types
x <- seq(-4, 8, by = 0.1)

# Define all function types
linear_func <- 0.5 * x + 2
quadratic_func <- 0.2 * (x - 2)^2 + 1
polynomial_func <- 0.02 * x^3 - 0.1 * x^2 + x + 3
exponential_func <- 2^(0.5 * x)
logarithmic_func <- ifelse(x > 0, 2 * log(x) + 5, NA)
trigonometric_func <- 3 * sin(x) + 5

# Create the plot
plot(x, linear_func, type = "l", lwd = 3, col = "#007bff", 
     ylim = c(-5, 15), xlim = c(-4, 8),
     main = "Comparison of Function Types: Different Relations Between Variables",
     xlab = "Independent Variable (x)", ylab = "Dependent Variable f(x)",
     cex.main = 1.2, cex.lab = 1.1)

# Add all other functions
lines(x, quadratic_func, lwd = 3, col = "#6f42c1")
lines(x, polynomial_func, lwd = 3, col = "#fd7e14")
lines(x, exponential_func, lwd = 3, col = "#dc3545")
lines(x[x > 0], logarithmic_func[x > 0], lwd = 3, col = "#20c997")
lines(x, trigonometric_func, lwd = 3, col = "#6c757d")

# Add grid for better readability
grid(col = "lightgray", lty = 2)

# Add comprehensive legend
legend("topleft", 
       legend = c("Linear: f(x) = 0.5x + 2", 
                  "Quadratic: f(x) = 0.2(x-2)² + 1",
                  "Polynomial: f(x) = 0.02x³ - 0.1x² + x + 3",
                  "Exponential: f(x) = 2^(0.5x)",
                  "Logarithmic: f(x) = 2ln(x) + 5",
                  "Trigonometric: f(x) = 3sin(x) + 5"),
       col = c("#007bff", "#6f42c1", "#fd7e14", "#dc3545", "#20c997", "#6c757d"),
       lwd = 3, cex = 0.9, bg = "white")

# Add annotations to highlight key differences
text(-2, 12, "Different functions capture\ndifferent types of relationships:", 
     cex = 1.1, font = 2, adj = 0)
text(-2, 10, "• Linear: Constant rate of change", cex = 0.9, adj = 0)
text(-2, 9.2, "• Quadratic: Accelerating change", cex = 0.9, adj = 0)
text(-2, 8.4, "• Exponential: Explosive growth", cex = 0.9, adj = 0)
text(-2, 7.6, "• Logarithmic: Diminishing growth", cex = 0.9, adj = 0)
text(-2, 6.8, "• Trigonometric: Periodic patterns", cex = 0.9, adj = 0)
text(-2, 6.0, "• Polynomial: Complex curves", cex = 0.9, adj = 0)

Comparison of Common Function Types over [-4, 8]

Composit Function and Inverse Function

We have two functions: \(g,f: \mathbb{R}\to \mathbb{R}\), \(z = g(y)\) and \(y = f(x)\).

\(z=g(y)=g(f(x))\) notation \(z = (g\circ f)(x)\)

Example: exchange rate: 1 AUD = 0.6 USD, 1 USD = 150 YEN

\(Y_{USD} = 0.6 X_{AUD}\), \(Y_{USD} = f(X_{AUD})\)

\(Z_{YEN} = 150 Y_{USD}\) \(Z_{YEN} = g(Y_{USD})\)

\(Z_{YEN} = (g\circ f )(X_{AUD}) = g(f(X_{AUD}))=150*(0.6*X_{AUD}) = 90X_{AUD}\)

Inverse Function

If \(g(f(x))=x\) then \(g\) is called the inverse function of \(f\) and denoted as \(f^{-1}\).

\(Y_{USD} = 0.6 X_{AUD}\), \(Y_{USD} = f(X_{AUD})\)

1 USD = 1/0.6 AUD = 1.6666 AUD

\(Z_{AUD} = 1/0.6 Y_{USD}\) \(Z_{AUD} = g(Y_{USD})\)

Sumation Notation as a specific function

Calculate the following without using computer

1. Calculate without using computer the sum of \(\sum_{i=1}^{10} i\) and Write R code to complete this task.

\[\sum_{i=1}^{10} i = 1+2+3+4+5+6+7+8+9+10 = 1+9 + 2+8 + 3+7 + 4+6 +5 = 55\]

y = 0
for (i in 1:10) y = y + i
y

## [1] 55

Sum <- function(n){
  y=0
  for (i in 1:n) y = y + i
return(y)
}

Sum(10)

## [1] 55

2. Calculate without using computer the sum of \(\sum_{i=0}^n r^i\) and write R code to complete this task. \[S_n = \sum_{i=0}^n r^i\] \[rS_n = \sum_{i=1}^{n+1} r^i\] \[S_n-rS_n = \sum_{i=0}^{n} r^i - \sum_{i=1}^{n+1} r^i = r^0-r^{n+1}=S_n(1-r)\] \[S_n = \frac{r^0-r^{n+1}}{1-r}\]

Sum_r <- function(n,r){
  y=0
  for (i in 0:n) y = y + r^i
return(y)
}
r = 0.6
n = 10
Sum_r(n=10,r = 0.6)

## [1] 2.49093

### compare with the formula

(1-0.6^{11})/(1-0.6)

## [1] 2.49093

Sample Code Demonstrations

Experience hands-on learning with interactive R code! Modify the code, experiment with different values, and see immediate results:

Interactive Quiz 2.1

Interactive Quiz 2.2

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Knowledge Point 3: Functions of More Than One Variable

Verbal Scenario: The Investment Portfolio Manager

Consider an investment portfolio manager who needs to understand how returns depend on multiple factors:

3.1 The Revenue Analysis

A company’s revenue depends on both price and quantity: R(P, Q) = P × Q - P = price per unit - Q = quantity sold - Shows how two variables interact to determine total revenue

3.2 The Portfolio Return

An investment portfolio with three stocks has return: R = w₁R₁ + w₂R₂ + w₃R₃ - w₁, w₂, w₃ = portfolio weights (must sum to 1) - R₁, R₂, R₃ = individual stock returns - Shows linear combination of multiple variables

This demonstrates multivariable functions where outputs depend on multiple inputs simultaneously.

Formal Presentation

Definition 3.1 (Multivariable Function): A function of several variables is a rule that assigns to each ordered set of input values exactly one output value. We write f(x₁, x₂, …, xₙ) = y.

Key Concepts: 1. Domain: Set of all possible input combinations 2. Partial dependence: Output changes when one variable changes while others remain constant 3. Interaction effects: Variables may influence each other’s impact on the output

Common Multivariable Functions

Two Variables Case

1. Revenue Function: R(P, Q) = P × Q
   • Price and quantity determine total revenue
   • Shows multiplicative relationship between variables
2. Production Function: Q(L, K) = ALᵅKᵝ (Cobb-Douglas)
   • Labor (L) and Capital (K) produce output (Q)
   • Shows how inputs combine to create output

Three Variables Case

1. Portfolio Return: R = w₁R₁ + w₂R₂ + w₃R₃
   • Three stock weights and returns determine portfolio return
   • Linear combination with constraint: w₁ + w₂ + w₃ = 1
2. Cost Function: C(Q, w, r) = wL + rK + FC
   • Quantity, wage rate, and rental rate determine total cost
   • Shows how multiple cost factors combine

Economic Functions with Concrete Examples

Production Functions

General Form: Q(L, K) = ALᵅKᵝ

Concrete Example: Q(L, K) = 10L^0.6K0.4 - A = 10 (technology parameter) - α = 0.6 (labor elasticity) - β = 0.4 (capital elasticity) - Returns to scale: α + β = 1 (constant returns)

Utility Functions

General Form: U(x, y) = xᵅy^(1-α)

Concrete Example: U(x, y) = x^0.5y0.5 - Equal preference weights for both goods - Represents perfect substitutes with diminishing marginal utility - Indifference curves are rectangular hyperbolas

Graphical Presentation of Function of Multiple Variables

Generally we have multiple ways to illustrate a function of more than one variable:

• Surface Plot
• Isoquants
• Heatmap
• Slice line _ gradian field

Rise field and Isoquants

Illustration of Cobb-Douglas Production of Function

Graphical Presentation of Multivariables Function

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Knowledge Point 4: Functions and Equations - Solving for Roots

Verbal Scenario: The Business Analyst

A business analyst encounters various equation-solving challenges:

4.1 The Break-Even Analysis

A startup company wants to find its break-even point where total revenue equals total cost:

• Revenue Function: R(Q) = 25Q (selling price $25 per unit)
• Cost Function: C(Q) = 5000 + 15Q (fixed costs $5000, variable cost $15 per unit)

To find break-even: Set R(Q) = C(Q) 25Q = 5000 + 15Q 10Q = 5000 Q = 500 units

4.2 The Market Equilibrium

Finding where supply equals demand:

• Supply Function: Qs = -100 + 2P (suppliers willing to sell)
• Demand Function: Qd = 300 - P (consumers willing to buy)

At equilibrium: Qs = Qd -100 + 2P = 300 - P 3P = 400 P = $133.33, Q = 166.67 units

These are examples of solving equations to find where functions intersect.

Formal Presentation

Definition 4.1 (Equation): An equation is a mathematical statement that two expressions are equal. When one side is zero, we call the solutions roots or zeros.

Definition 4.2 (Root/Zero): A root of f(x) = 0 is a value x = r such that f(r) = 0.

Types of Equations and Solution Methods:

1. Linear Equations: ax + b = 0

• Solution: x = -b/a
• Always has exactly one solution (if a ≠ 0)

Concrete Example: 3x - 9 = 0 - Solution: x = 9/3 = 3 - Graphical interpretation: The line y = 3x - 9 crosses the x-axis (y = 0) at x = 3

2. Quadratic Equations: ax² + bx + c = 0

• Quadratic formula: x = (-b ± √(b² - 4ac))/(2a)
• Can have 0, 1, or 2 real solutions

3. Higher-order Polynomial Equations: aₙxⁿ + … + a₁x + a₀ = 0

• May require numerical methods for n > 4
• Can have up to n real roots

Concrete Example: Fourth-order polynomial with 4 roots - Factored form: y = (x + 2)(x - 1)(x - 3)(x - 5) - Expanded: y = x⁴ - 7x³ + 7x² + 27x - 30 - Four roots: x = -2, 1, 3, 5

4. Transcendental Equations: Involving exponential, logarithmic, or trigonometric functions

• Usually require numerical methods

Concrete Example: eˣ - 3 = 0 - Solution: x = ln(3) ≈ 1.099 - Graphical interpretation: The curve y = eˣ - 3 crosses the x-axis at x = ln(3)

Economic Applications

Break-Even Analysis

Mathematical Formulation: - Revenue: R(Q) = 25Q - Cost: C(Q) = 5000 + 15Q - Break-even condition: R(Q) = C(Q)

Solution: 25Q = 5000 + 15Q 10Q = 5000 Q = 500 units

At break-even: Revenue = Cost = $12,500

Market Equilibrium: Supply = Demand

Mathematical Formulation: - Supply: Qs = -100 + 2P - Demand: Qd = 300 - P - Equilibrium condition: Qs = Qd

Solution: -100 + 2P = 300 - P 3P = 400 P = $133.33

Substituting back: Q = -100 + 2(133.33) = 166.67 units

Solving an equation and equation system

Find the roots

Solving an equation system

Interactive Quiz 4.1

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Comprehensive Review Questions

This section provides review and application questions designed to assess understanding at five different levels. Each level builds upon the previous one, fostering deeper comprehension and the ability to apply concepts in increasingly complex situations.

Level 1: Descriptive Understanding (Basic Concepts)

This level tests your ability to recognize, define, and describe fundamental concepts.

Question 1.1

Define what a function is and give three examples from everyday life.

Sample Answer: A function is a rule that assigns to each input exactly one output. Examples: (1) The cost of gasoline depends on the number of gallons purchased, (2) Your age depends on the current date, (3) The area of a circle depends on its radius.

Grading Criteria (10 points): - Correct definition (4 points) - Three appropriate examples (6 points, 2 each)

Question 1.2

What is the difference between a variable and a constant? Provide examples of each.

Sample Answer: A variable is a symbol that can take on different values (like x, y, t), while a constant is a fixed value that doesn’t change (like π = 3.14159, or the number 5 in the expression 5x + 3).

Grading Criteria (8 points): - Correct definition of variable (3 points) - Correct definition of constant (3 points) - Appropriate examples (2 points)

Question 1.3

List and briefly describe four different ways to represent a function.

Sample Answer: (1) Table: showing input-output pairs in rows and columns, (2) Graph: visual plot showing the relationship, (3) Formula: algebraic expression like f(x) = 2x + 3, (4) Code: computational algorithm or program.

Grading Criteria (12 points): - Four representations listed (4 points) - Brief descriptions (8 points, 2 each)

Question 1.4

What does it mean for a function to have “domain” and “range”? Give an example.

Sample Answer: Domain is the set of all possible input values, and range is the set of all possible output values. For f(x) = √x, the domain is x ≥ 0 (can’t take square root of negative numbers), and the range is y ≥ 0 (square roots are non-negative).

Grading Criteria (10 points): - Correct definition of domain (3 points) - Correct definition of range (3 points) - Appropriate example with correct domain and range (4 points)

Question 1.5

Explain the difference between a linear function and a quadratic function.

Sample Answer: A linear function has the form f(x) = mx + b and graphs as a straight line. A quadratic function has the form f(x) = ax² + bx + c and graphs as a parabola (U-shaped curve).

Grading Criteria (8 points): - Correct form of linear function (2 points) - Correct graph description for linear (2 points) - Correct form of quadratic function (2 points) - Correct graph description for quadratic (2 points)

Question 1.6

What does it mean to “solve an equation”? What are the “roots” of a function?

Sample Answer: Solving an equation means finding the value(s) of the variable that make the equation true. The roots (or zeros) of a function are the input values where the function equals zero, i.e., where f(x) = 0.

Grading Criteria (8 points): - Correct definition of solving an equation (4 points) - Correct definition of roots/zeros (4 points)

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Level 2: Operational Understanding (Calculation and Practice)

This level tests your ability to perform calculations and apply procedures correctly.

Question 2.1

Given f(x) = 3x² - 2x + 5, calculate: a) f(2) b) f(-1) c) f(0)

Sample Answer: a) f(2) = 3(2)² - 2(2) + 5 = 3(4) - 4 + 5 = 12 - 4 + 5 = 13 b) f(-1) = 3(-1)² - 2(-1) + 5 = 3(1) + 2 + 5 = 10 c) f(0) = 3(0)² - 2(0) + 5 = 5

Grading Criteria (12 points): - Part (a): 4 points for correct calculation - Part (b): 4 points for correct calculation - Part (c): 4 points for correct calculation

Question 2.2

Solve the following equations: a) 4x - 12 = 0 b) x² - 5x + 6 = 0 c) 2x + 7 = 3x - 5

Sample Answer: a) 4x = 12, so x = 3 b) Using factoring: (x - 2)(x - 3) = 0, so x = 2 or x = 3 c) 2x - 3x = -5 - 7, so -x = -12, therefore x = 12

Grading Criteria (15 points): - Part (a): 3 points for correct solution - Part (b): 6 points (3 for method, 3 for correct roots) - Part (c): 6 points for correct solution

Question 2.3

A company’s cost function is C(q) = 500 + 25q, where q is the quantity produced. Calculate the total cost for producing 20, 50, and 100 units.

Sample Answer: - C(20) = 500 + 25(20) = 500 + 500 = $1000 - C(50) = 500 + 25(50) = 500 + 1250 = $1750 - C(100) = 500 + 25(100) = 500 + 2500 = $3000

Grading Criteria (12 points): - Each calculation: 4 points (2 for setup, 2 for correct answer)

Question 2.4

For the function g(x, y) = 2x + 3y - 1, calculate: a) g(1, 2) b) g(-2, 4) c) g(0, 0)

Sample Answer: a) g(1, 2) = 2(1) + 3(2) - 1 = 2 + 6 - 1 = 7 b) g(-2, 4) = 2(-2) + 3(4) - 1 = -4 + 12 - 1 = 7 c) g(0, 0) = 2(0) + 3(0) - 1 = -1

Grading Criteria (12 points): - Each calculation: 4 points for correct answer

Question 2.5

Find the break-even point for a business where Revenue = 40q and Cost = 800 + 15q, where q is the quantity sold.

Sample Answer: Set Revenue = Cost: 40q = 800 + 15q 40q - 15q = 800 25q = 800 q = 32 units

Grading Criteria (10 points): - Correct equation setup (4 points) - Correct algebraic manipulation (4 points) - Correct final answer (2 points)

Question 2.6

Use the quadratic formula to solve 2x² + 3x - 2 = 0.

Sample Answer: Using x = (-b ± √(b² - 4ac))/(2a) with a = 2, b = 3, c = -2: x = (-3 ± √(9 - 4(2)(-2)))/(2(2)) x = (-3 ± √(9 + 16))/4 x = (-3 ± √25)/4 x = (-3 ± 5)/4 So x = 2/4 = 0.5 or x = -8/4 = -2

Grading Criteria (15 points): - Correct formula identification (3 points) - Correct substitution (4 points) - Correct discriminant calculation (4 points) - Correct final answers (4 points)

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Level 3: Causal Understanding (Reasoning and Explanation)

This level tests your ability to explain why things work and understand cause-and-effect relationships.

Question 3.1

Explain why the function f(x) = 1/x is undefined at x = 0. What does this tell us about the domain of this function?

Sample Answer: The function f(x) = 1/x is undefined at x = 0 because division by zero is undefined in mathematics. When we try to evaluate 1/0, there’s no finite number that, when multiplied by 0, gives 1. This tells us that the domain of f(x) = 1/x is all real numbers except 0, written as (-∞, 0) ∪ (0, ∞).

Grading Criteria (12 points): - Explanation of why division by zero is undefined (6 points) - Correct identification of domain (4 points) - Clear mathematical reasoning (2 points)

Question 3.2

Why do quadratic functions always have either 0, 1, or 2 real roots? Explain using the discriminant.

Sample Answer: The number of real roots depends on the discriminant Δ = b² - 4ac. When Δ > 0, the square root in the quadratic formula gives a real number, leading to two distinct real roots. When Δ = 0, the square root is zero, giving exactly one real root (a repeated root). When Δ < 0, the square root of a negative number is not real, so there are no real roots (but two complex roots).

Grading Criteria (15 points): - Correct explanation for Δ > 0 (5 points) - Correct explanation for Δ = 0 (5 points) - Correct explanation for Δ < 0 (5 points)

Question 3.3

A production function shows decreasing marginal returns. Explain what this means and why it typically occurs in real-world production.

Sample Answer: Decreasing marginal returns means that as you add more of one input (like labor), each additional unit produces less additional output than the previous unit. This occurs because other inputs (like machinery or workspace) become limiting factors. For example, adding the 10th worker to a small factory might increase output by 5 units, but adding the 20th worker might only increase output by 2 units because workers start getting in each other’s way.

Grading Criteria (15 points): - Correct definition of decreasing marginal returns (6 points) - Explanation of why it occurs (6 points) - Appropriate real-world example (3 points)

Question 3.4

Explain why exponential functions are appropriate for modeling population growth, while linear functions are not.

Sample Answer: Exponential functions are appropriate because population growth depends on the current population size - larger populations produce more offspring. If a population grows by 5% per year, a population of 1000 adds 50 individuals, while a population of 10,000 adds 500 individuals. Linear functions assume constant absolute growth regardless of current size, which doesn’t reflect biological reality where reproductive capacity increases with population size.

Grading Criteria (15 points): - Explanation of why exponential fits population growth (8 points) - Explanation of why linear doesn’t fit (4 points) - Clear reasoning about proportional vs. absolute growth (3 points)

Question 3.5

Why might a cost function be quadratic rather than linear? Give a specific business example.

Sample Answer: Cost functions can be quadratic when there are economies or diseconomies of scale. For example, a manufacturing company might have increasing marginal costs due to overtime wages, equipment strain, or space constraints. The cost function C(q) = 1000 + 50q + 0.1q² shows fixed costs ($1000), constant variable costs ($50 per unit), plus increasing costs (0.1q²) that represent inefficiencies at high production levels, such as paying overtime rates or using less efficient backup equipment.

Grading Criteria (15 points): - Explanation of economies/diseconomies of scale (6 points) - Specific business example (6 points) - Connection between quadratic form and real-world factors (3 points)

Question 3.6

Explain why graphical methods for solving equations can be useful even when algebraic methods exist.

Sample Answer: Graphical methods provide visual insight into the behavior of functions and can reveal information that algebraic methods might miss. They show how many solutions exist, whether solutions are approximate or exact, and help identify the general behavior of the function. For complex equations where algebraic solutions are difficult or impossible, graphical methods can provide good approximations. They also help verify algebraic solutions and can reveal errors in calculations.

Grading Criteria (12 points): - Visual insight explanation (4 points) - Approximation capabilities (4 points) - Verification and error-checking benefits (4 points)

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Level 4: Predictive Understanding (Application and Extension)

This level tests your ability to apply concepts to new situations and make predictions.

Question 4.1

A tech startup’s user growth follows the function U(t) = 1000e^(0.15t), where t is time in months. Predict the number of users after 12 months and explain what assumptions this model makes.

Sample Answer: U(12) = 1000e^(0.15×12) = 1000e^1.8 ≈ 1000 × 6.05 ≈ 6,050 users

Assumptions: (1) Growth rate remains constant at 15% per month, (2) No external factors limit growth, (3) No competition effects, (4) Unlimited market size, (5) No user churn or abandonment.

Grading Criteria (18 points): - Correct calculation (8 points) - Identification of key assumptions (10 points, 2 each)

Question 4.2

Design a pricing strategy for a product where demand follows D(p) = 500 - 10p and production cost is C(q) = 2000 + 15q. What price maximizes profit?

Sample Answer: Revenue: R = p × D(p) = p(500 - 10p) = 500p - 10p² Since q = D(p), Cost becomes: C = 2000 + 15(500 - 10p) = 2000 + 7500 - 150p = 9500 - 150p Profit: π = R - C = (500p - 10p²) - (9500 - 150p) = 650p - 10p² - 9500

To maximize, take derivative and set to zero: dπ/dp = 650 - 20p = 0 p = 32.5

At p = $32.50: D(32.5) = 500 - 10(32.5) = 175 units Maximum profit = 650(32.5) - 10(32.5)² - 9500 = $1,062.50

Grading Criteria (25 points): - Correct revenue function (5 points) - Correct cost function transformation (5 points) - Correct profit function (5 points) - Correct optimization method (5 points) - Correct final answer (5 points)

Question 4.3

A pharmaceutical company models drug concentration in blood as C(t) = 20te^(-0.5t), where t is hours after injection. When will concentration be highest, and what factors might make this model inaccurate?

Sample Answer: To find maximum, take derivative: C’(t) = 20e^(-0.5t) + 20t(-0.5)e^(-0.5t) = 20e^(-0.5t)(1 - 0.5t) Set C’(t) = 0: 1 - 0.5t = 0, so t = 2 hours

Maximum concentration occurs at t = 2 hours: C(2) = 20(2)e^(-1) ≈ 14.7 units

Factors affecting accuracy: (1) Individual metabolism differences, (2) Drug interactions, (3) Kidney/liver function variations, (4) Age and weight differences, (5) Multiple dosing effects, (6) Food interactions.

Grading Criteria (20 points): - Correct derivative calculation (6 points) - Correct time of maximum (4 points) - Correct maximum value (4 points) - Realistic limiting factors (6 points)

Question 4.4

An environmental scientist models pollution cleanup as P(t) = P₀e^(-kt), where P₀ = 1000 units and k = 0.1 per year. How long until pollution drops to 10% of original level? What if cleanup efforts double (k = 0.2)?

Sample Answer: For 10% of original: 0.1P₀ = P₀e^(-0.1t) 0.1 = e^(-0.1t) ln(0.1) = -0.1t t = -ln(0.1)/0.1 = 2.303/0.1 ≈ 23.0 years

With doubled efforts (k = 0.2): 0.1 = e^(-0.2t) t = -ln(0.1)/0.2 ≈ 11.5 years

Doubling cleanup efforts halves the time needed, demonstrating the exponential relationship between effort and results.

Grading Criteria (18 points): - Correct equation setup (4 points) - Correct solution for k = 0.1 (6 points) - Correct solution for k = 0.2 (6 points) - Insight about relationship (2 points)

Question 4.5

A city’s water usage follows W(T, P) = 50 + 2T + 0.001P², where T is temperature (°C) and P is population. If temperature increases by 5°C and population grows by 20%, predict the change in water usage for a city with current T = 25°C and P = 100,000.

Sample Answer: Current usage: W(25, 100000) = 50 + 2(25) + 0.001(100000)² = 50 + 50 + 10000 = 10,100 units

New conditions: T = 30°C, P = 120,000 New usage: W(30, 120000) = 50 + 2(30) + 0.001(120000)² = 50 + 60 + 14400 = 14,510 units

Change: 14,510 - 10,100 = 4,410 units (43.7% increase)

The population growth has a much larger impact than temperature due to the quadratic relationship.

Grading Criteria (20 points): - Correct current usage calculation (6 points) - Correct new usage calculation (8 points) - Correct change calculation (4 points) - Insight about relative impacts (2 points)

Question 4.6

A renewable energy company’s profit depends on government subsidies (S) and oil prices (O): π(S, O) = -1000 + 50S + 30O - 0.1S². If subsidies might be cut from $20 to $10 per unit, what oil price would maintain current profit levels?

Sample Answer: Current profit with S = 20: π(20, O) = -1000 + 50(20) + 30O - 0.1(20)² = -1000 + 1000 + 30O - 40 = 30O - 40

With reduced subsidies S = 10: π(10, O) = -1000 + 50(10) + 30O - 0.1(10)² = -1000 + 500 + 30O - 10 = 30O - 510

To maintain same profit: 30O - 40 = 30O_new - 510 This gives: O_new = O + 470/30 ≈ O + 15.67

Oil prices would need to increase by approximately $15.67 per barrel to compensate for the subsidy reduction.

Grading Criteria (22 points): - Correct current profit function (6 points) - Correct new profit function (6 points) - Correct equation setup (5 points) - Correct solution (5 points)

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Level 5: Empathic Understanding (Explaining to Others)

This level tests your ability to teach concepts to others and communicate mathematical ideas clearly.

Question 5.1

You’re tutoring a high school student who asks: “Why do we need to learn about functions? When will I ever use this?” Write a clear, encouraging response with concrete examples.

Sample Answer: “Great question! Functions are everywhere in daily life, even if we don’t always recognize them. When you use GPS, it’s calculating functions to find the shortest route. When you post on social media, algorithms use functions to decide what appears in your feed. If you ever want to budget money, start a business, or understand how your car’s fuel efficiency works, you’re using functions.

Think of functions as ‘cause and effect’ relationships: How does the amount you study affect your test scores? How does the temperature outside affect your heating bill? How does the number of hours you work affect your paycheck? Once you understand these patterns, you can make better decisions and predictions.

Functions give you a superpower: the ability to see patterns and make predictions about the world around you. That’s incredibly valuable in any career path you choose!”

Grading Criteria (20 points): - Addresses student’s concern directly (4 points) - Provides concrete, relatable examples (8 points) - Encouraging and positive tone (4 points) - Clear explanation of value/relevance (4 points)

Question 5.2

Create a simple analogy to explain the concept of domain and range to someone who has never studied mathematics formally.

Sample Answer: “Think of a function like a vending machine. The domain is like all the button numbers you can press - maybe buttons 1 through 50. You can’t press button 51 because it doesn’t exist, just like some functions can’t accept certain input values.

The range is like all the different snacks that might actually come out. Even though there are 50 buttons, maybe only 20 different types of snacks are loaded in the machine. So while you can press 50 different buttons (domain), you’ll only get 20 different results (range).

For example, if f(x) = x², the domain might be all real numbers (you can square any number), but the range is only non-negative numbers (you can’t get a negative result when squaring real numbers), just like how our vending machine can’t give you a snack that isn’t loaded inside it.”

Grading Criteria (18 points): - Clear, accessible analogy (8 points) - Correct explanation of domain (4 points) - Correct explanation of range (4 points) - Mathematical example that connects to analogy (2 points)

Question 5.3

A business owner says: “I don’t understand why my cost function is curved instead of a straight line. Shouldn’t costs just increase steadily with production?” Explain this clearly.

Sample Answer: “That’s a very logical question! You’re thinking of a straight line because you’re imagining that each additional unit costs the same to produce. And sometimes that’s true for small increases in production.

But think about what happens in your business as you produce more and more. At first, you might get more efficient - buying materials in bulk gets you discounts, and your workers get better at their jobs. This could actually make each additional unit cheaper (costs curve downward).

But eventually, you hit limits. Maybe you need to pay overtime wages, or use older, less efficient equipment, or rent additional space at higher rates. Now each additional unit becomes more expensive (costs curve upward).

It’s like driving a car - at first, going faster might improve your fuel efficiency, but eventually, going too fast makes your car work harder and use more gas per mile. The curved line captures these real-world complexities that a straight line misses.”

Grading Criteria (20 points): - Acknowledges the logical nature of the question (3 points) - Explains economies of scale (5 points) - Explains diseconomies of scale (5 points) - Uses relatable analogy (4 points) - Clear, non-technical language (3 points)

Question 5.4

Design a 5-minute lesson plan to teach the concept of solving equations to middle school students, including an activity.

Sample Answer: Lesson Plan: “Equation Detective” (5 minutes)

Minute 1: Hook “You’re detectives solving a mystery! Someone stole cookies from the jar. We know: ‘Someone took some cookies, and now there are 3 left. Originally there were 12 cookies. How many were stolen?’”

Minutes 2-3: Concept Introduction “This is an equation: 12 - x = 3, where x is the unknown (stolen cookies). Solving equations means finding the mystery number. We use ‘opposite operations’ - if someone subtracted, we add back.”

Minutes 3-4: Guided Practice Activity Students work in pairs with balance scales (or drawings). “If 12 - x = 3, add x to both sides: 12 = 3 + x. Now subtract 3 from both sides: 9 = x.” Use physical objects to show balance.

Minute 5: Wrap-up “Quick check: If 9 cookies were stolen from 12, how many are left? 12 - 9 = 3 ✓ The detective solved the case!”

Grading Criteria (25 points): - Engaging hook/motivation (5 points) - Clear concept explanation (5 points) - Hands-on activity (5 points) - Age-appropriate language (5 points) - Realistic timing and structure (5 points)

Question 5.5

Write an email to a parent explaining why their child is struggling with multivariable functions and suggesting specific ways to help at home.

Sample Answer: Subject: Supporting [Child’s Name] with Multivariable Functions

Dear [Parent’s Name],

I wanted to reach out about [Child’s Name]’s progress with multivariable functions. This is one of the more challenging topics because it requires thinking about multiple changing factors simultaneously - like how both study time AND sleep quality affect test performance.

Many students struggle here because they’re used to simpler “one cause, one effect” thinking. [Child’s Name] is actually doing well with the calculations but needs more practice visualizing these relationships.

Here are some ways you can help at home:

1. Real-world connections: Discuss family decisions that depend on multiple factors. “How do we decide where to go on vacation?” (cost, weather, distance, activities available)

2. Cooking analogies: “How does the taste of cookies depend on both baking time AND temperature?” This makes abstract math concrete.

3. Encourage questions: If [Child’s Name] says “This doesn’t make sense,” ask them to explain what they DO understand first. Often they know more than they think.

4. Practice patience: This topic requires mental flexibility that develops over time. Celebrate small victories!

Please don’t hesitate to contact me if you have questions. [Child’s Name] is making good progress, and with continued practice, these concepts will click.

Best regards, [Teacher’s Name]

Grading Criteria (25 points): - Professional, supportive tone (5 points) - Clear explanation of the difficulty (5 points) - Specific, actionable suggestions (10 points) - Encouragement and realistic expectations (5 points)

Question 5.6

Create a visual metaphor or story that would help someone understand why we use different types of functions (linear, quadratic, exponential) for different real-world situations.

Sample Answer: “The Three Runners Story”

Imagine three friends running different races:

Linear Larry runs at a perfectly steady pace. Every minute, he covers exactly the same distance. His progress looks like a straight line going up - steady, predictable, reliable. This is like your cell phone bill: $30 base cost plus $10 for each gigabyte you use. Linear functions work when things change at a constant rate.

Quadratic Quinn starts slow but speeds up as she gets warmed up (or starts fast but gets tired). Her distance covered each minute keeps changing - maybe 1 mile, then 2 miles, then 3 miles in successive minutes. Her total distance curves upward like a smile. This is like the area of a square: double the side length, and the area goes up by four times (2² = 4). Quadratic functions work when the rate of change itself is changing.

Exponential Emma is amazing - she doubles her speed every minute! First minute: 1 mph. Second minute: 2 mph. Third minute: 4 mph. Her progress shoots up like a rocket. This is like a viral video: first day 100 views, second day 200, third day 400. Exponential functions work when growth depends on how big something already is.

Each runner is perfect for their own type of race, just like each function type is perfect for its own type of real-world situation!

Grading Criteria (25 points): - Creative, memorable metaphor (8 points) - Accurate representation of linear functions (5 points) - Accurate representation of quadratic functions (6 points) - Accurate representation of exponential functions (6 points)

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Summary and Reflection

Key Takeaways

Today we’ve explored the fundamental building blocks of calculus:

1. Numbers and Variables: The basic language of mathematics
2. Functions: Mathematical tools for describing relationships
3. Multiple Representations: Tables, graphs, formulas, and code
4. Multivariable Functions: Real-world complexity with multiple inputs
5. Real-World Applications: How mathematics describes our world
6. Equation Solving: Finding specific values that satisfy conditions

Metacognitive Reflection Questions

Before moving to the next section, reflect on these questions:

1. Which representation of functions (table, graph, formula, code) feels most natural to you? Why?
2. Can you think of a relationship in your own life that would be best modeled by each type of function we discussed?
3. What was the most challenging concept today, and what strategy helped you understand it?
4. How has your understanding of “functions” changed from the beginning to the end of this lecture?

Looking Ahead

In our next section, we’ll dive deeper into functions and their graphs, exploring how to analyze their behavior, find their key features, and use them to solve more complex problems. The foundation we’ve built today will be essential for understanding limits, continuity, and eventually derivatives.

Remember: Mathematics is a language for describing patterns in our world. The more fluent you become in this language, the more clearly you’ll see the hidden structures that govern everything from economics to physics to biology.

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End of Section 1 Lecture Notes